Chapter 5 Possibilities Chapter 5 Possibilities and Probability and Probability Counting Permutations Combinations Probability
Chapter 5 Possibilities and Chapter 5 Possibilities and ProbabilityProbability
CountingPermutationsCombinationsProbability
5.1 Counting5.1 Counting
Example 5.1: Ice cream cones
Flavor: chocolate , vanilla, strawberry
Cones: sugar, regular
How many different varieties?
Tree diagramTree diagram
sugar choc reg sugar van reg sugar str reg
3*2=6 choices
Rule 1: Multiplication RuleRule 1: Multiplication Rule
A choice consists of 2 distinct steps1st step can be made in m different waysFor each of these, 2nd step can be made in n
different ways
Then the whole choice can be made in m*n ways
Example 5.2Example 5.2
4 horses in a race How many ways can we pick a first
and second place finisher?
Answer: 43=12
You can get it from a tree diagram as well
Choice of 1st place choices of 2nd place
4*3=12 ways
A
B
C
DA
B
C
ABD
AC
D
B
C
D
Rule 2: Generalized Multiplication Rule 2: Generalized Multiplication RuleRule
A choice consists of k steps;
Step 1 can be made in n1 ways;
Step 2 can be made in n2 ways;
… …
Step k can be made in nk ways ;
Then whole choice can be made in n1n2…nk ways
Example 5.3Example 5.3
5 horses in a raceHow many ways can we pick a 1st , 2nd, and
3rd place finisher?
Example 5.4Example 5.4
A multiple choice exam has 5 questionsEach question has 4 possible answers
What is the number of ways to answer the exam?
Example 5.5Example 5.5
How many license plates can be formed with 3 letters followed by 3 numbers?
Example 5.6Example 5.6
How many ways can we write the letters
O W L ?
5.2 Permutations5.2 Permutations
The number of ways to order r of n objects
nPr=n(n-1)(n-2)…(n-r+1)
(application of multiplication rule)
Example 5.7Example 5.7
How many ways can we put 3 cards from a deck of cards in order?
52 51 50
(52)(51)(50)=132,600
Notation: n!=(1)(2)Notation: n!=(1)(2)……(n-1)(n)(n-1)(n)
The number of ways to order (permute) n of n objects is
nPn = n(n-1)(n-2)…(1) = n!
n! is called n-factorial
3!=(1)(2)(3)=6
4!=(1)(2)(3)(4)=3!(4)=24
5!=4!(5)=120
(0!=1)
Example 5.8Example 5.8
How many ways can we order 5 horses?
5!=120
Example 5.9Example 5.9
How many ways can we arrange the letters OLWS?
The Number of PermutationsThe Number of Permutations
)!(
!)1()1(
rn
nrnnnPrn
60)3)(4)(5()1)(2(
)1)(2)(3)(4)(5(
)!35(
!535
P
ExerciseExercise
How many different ways can we inject 3 of 15 mice with 3 different doses of a serum?
Exercise 5.21Exercise 5.21
In optics kits there are 5 concave lenses, 5 convex lenses, 2 prisms, and 3 mirrors. how many different ways can a person choose 1 of each kind?
Exercise 5.27Exercise 5.27
In how many different ways can a television director schedule a sponsor’s six different commercials during a telecast?
5.3 Combinations5.3 Combinations
How many ways can we choose 3 of 5 candidates to be in the final election?
Here the order of choosing the 3 finalists doesn’t matter!
Think this way Think this way …… ……
When we did worry about the order, there are (5)(4)(3)=60 ways
ABC ACB BAC BCA CAB CBA
ABD ADB BAD BDA DAB DBA
BCD BDC ...
… …
andand
Each choice of 3 candidates has 3!=6 ordered versions ABC ABC ACB BAC BCA CAB CBA
ThereforeTherefore
ABC ACB BAC BCA CAB CBAABC
ABD ADB BAD BDA DAB DBAABD
BCD BDC ... BCD
… …
The options are decreased by a factor of 6 compared to the ordered options
The number of ways to pick 3 out of 5The number of ways to pick 3 out of 5
10)1)(2(
)4)(5(
!3)!35(
!5
33
35
P
P
The Number of CombinationsThe Number of Combinations
The number of unordered ways to choose r of n objects is
(n choose r)
)!(!
!
rnr
n
r
nCrn
PascalPascal’’s Triangle s Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Example 5.10Example 5.10
Choose 5 cards from a deck
!5552
)1)(2)(3)(4)(5(
)48)(49)(50)(51)(52(
)!47)(1)(2)(3)(4)(5(
)!47)(48)(49)(50)(51)(52(
)!552(!5
!52552
P
Example 5.11Example 5.11
5 flavors of ice cream. Choose 2. Order doesn’t matter.
Example 5.12Example 5.12
6 candidates in a primary election. Choose 2 for a final election
ExerciseExercise
Calculate the number of ways in which a chain of ice cream stores can choose 2 of 12 locations for new franchises.
ExerciseExercise
A computer store carries 15 kinds of monitors. Calculate the number of ways in which we can purchase 3 different ones.
ExerciseExercise
In planning a garden we have 5 kinds of bushes to choose from and 10 kinds of flowers. How many ways can we choose 2 kinds of bushes and 4 kinds of flowers?
5.4 Probability5.4 Probability
In a deck of cards what is the probability of picking an ace?
What is meant by “Probability”?
Frequency interpretation: The probability of an event happening is the proportion of times that event would occur in the long run.
FactsFacts
Each card is equally likely to be chosen for a shuffled deck of cards
If there are “n” equally likely possibilities and “s” of these are a “success” , then the probability of a success is s/n
P(ace)= 4/52=1/13
P(red)=26/52=1/2
Example 5.13Example 5.13
Draw 2 cards from a deck. What is the probability that we get 2 aces?
Ideas Ideas
Each card is equally likely to be selected if one card is selected
All pairs of cards are equally likely to be selected if only two cards are selected
Any three cards are equally likely to be selected if only three cards are selected
… … (we call these randomness)
Solution to Example 5.13Solution to Example 5.13
# of possible ways to pick 2 cards
# of ways to pick 2 aces
Probability P(2 aces)
13262
)51)(52(
2
52
n
62
)3)(4(
2
4
s
0045.01326
6
Example 5.14Example 5.14
Pick up 3 cards, what is the probability of getting 2 aces and 1 king?
Let’s work it out!
Step 1
Step 2
Step 3 Probability P(2 aces and 1 king)=s/n= 24/22100
241
4
2
4
s
22100)1)(2)(3(
)50)(51)(52(
3
52
n
# of ways to pick 2 aces out of 4
# of ways to pick 1 king out of 4
Example 5.15Example 5.15
Pick up 5 cards. What is the probability of getting 3 aces?
Step 1
Step 2
Step 3 Probability
P(3 aces in 5 cards )=s/n=
4 48
3 2s
52
5n
# of ways to pick 3 aces out of 4
# of ways to pick other 2 cards
5
52
2
48
3
4
Example 5.16Example 5.16 Roll a red die and a white die. Find the probability that sum=3. red die: 1, 2, 3, 4, 5, 6 white die: 1, 2, 3, 4, 5, 6
n = # of outcomes= (ways for red to land)*(ways for white to land)= 6×6=36 pairs
s = # of ways sum=3(R=1, W=2), (R=2, W=1)
=2 Probability=2/36=1/18